Calculating $lim_{x to 0} frac{ln sin(x)}{lntan(x)}$
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I need some help calculating this limit:
$$lim_{x to 0} frac{ln sin(x)}{ln tan(x)}$$
limits
edited Nov 16 at 14:02
gimusi
85.5k74294
asked Nov 16 at 11:44
Lowie
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up vote
-1
down vote
favorite
I need some help calculating this limit:
$$lim_{x to 0} frac{ln sin(x)}{ln tan(x)}$$
limits
edited Nov 16 at 14:02
gimusi
85.5k74294
asked Nov 16 at 11:44
Lowie
162
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hint:As can be seen, the point is not in the domain of the function, so the limit does not exist.
– 1ENİGMA1
Nov 16 at 11:50
You can only take right hand limit. To find the value of the limit you simply have to apply L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 11:52
1
@1ENİGMA1 The existence of limits does not depend on whether the function is defined at the point. We only require that the point is a limit point of the domain.
– xbh
Nov 16 at 11:56
@xbh I know it. What I mean you can see answers.
– 1ENİGMA1
Nov 16 at 12:16
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I need some help calculating this limit:
$$lim_{x to 0} frac{ln sin(x)}{ln tan(x)}$$
limits
edited Nov 16 at 14:02
gimusi
85.5k74294
asked Nov 16 at 11:44
Lowie
162
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I need some help calculating this limit:
$$lim_{x to 0} frac{ln sin(x)}{ln tan(x)}$$
limits
limits
edited Nov 16 at 14:02
gimusi
85.5k74294
asked Nov 16 at 11:44
Lowie
162
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 14:02
gimusi
85.5k74294
asked Nov 16 at 11:44
Lowie
162
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 14:02
gimusi
85.5k74294
edited Nov 16 at 14:02
gimusi
85.5k74294
edited Nov 16 at 14:02
gimusi
85.5k74294
85.5k74294
asked Nov 16 at 11:44
Lowie
162
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 11:44
Lowie
162
asked Nov 16 at 11:44
Lowie
162
162
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hint:As can be seen, the point is not in the domain of the function, so the limit does not exist.
– 1ENİGMA1
Nov 16 at 11:50
You can only take right hand limit. To find the value of the limit you simply have to apply L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 11:52
1
@1ENİGMA1 The existence of limits does not depend on whether the function is defined at the point. We only require that the point is a limit point of the domain.
– xbh
Nov 16 at 11:56
@xbh I know it. What I mean you can see answers.
– 1ENİGMA1
Nov 16 at 12:16
add a comment |
Hint:As can be seen, the point is not in the domain of the function, so the limit does not exist.
– 1ENİGMA1
Nov 16 at 11:50
You can only take right hand limit. To find the value of the limit you simply have to apply L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 11:52
1
@1ENİGMA1 The existence of limits does not depend on whether the function is defined at the point. We only require that the point is a limit point of the domain.
– xbh
Nov 16 at 11:56
@xbh I know it. What I mean you can see answers.
– 1ENİGMA1
Nov 16 at 12:16
Hint:As can be seen, the point is not in the domain of the function, so the limit does not exist.
– 1ENİGMA1
Nov 16 at 11:50
Hint:As can be seen, the point is not in the domain of the function, so the limit does not exist.
– 1ENİGMA1
Nov 16 at 11:50
You can only take right hand limit. To find the value of the limit you simply have to apply L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 11:52
You can only take right hand limit. To find the value of the limit you simply have to apply L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 11:52
1
1
@1ENİGMA1 The existence of limits does not depend on whether the function is defined at the point. We only require that the point is a limit point of the domain.
– xbh
Nov 16 at 11:56
@1ENİGMA1 The existence of limits does not depend on whether the function is defined at the point. We only require that the point is a limit point of the domain.
– xbh
Nov 16 at 11:56
@xbh I know it. What I mean you can see answers.
– 1ENİGMA1
Nov 16 at 12:16
@xbh I know it. What I mean you can see answers.
– 1ENİGMA1
Nov 16 at 12:16
add a comment |
3 Answers
3
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up vote
1
down vote
HINT
As noticed we need $x>0$ then
$$frac{ln sin(x)}{ln tan(x)}=frac{ln frac{sin(x)}x+ln x}{ln frac{tan(x)}x+ln x}$$
answered Nov 16 at 12:07
gimusi
85.5k74294
add a comment |
up vote
1
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Just consider the reciprocal (for $x>0$):
$$frac{ln tan x}{ln sin x} = frac{ln sin x - ln cos x}{ln sin x} =1- frac{ln cos x}{ln sin x} stackrel{x to 0^+}{longrightarrow} 1$$
It follows $boxed{lim_{x to 0+} frac{ln sin(x)}{ln tan(x)} = 1}$.
answered Nov 16 at 12:17
trancelocation
8,0561519
add a comment |
up vote
0
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The domain of $ln x$ includes all positive values of $x$, so only the right-hand limit exists. If you want to find that, you can use L’Hôpital’s Rule.
$$lim_{x to 0} frac{ln(sin x)}{ln(tan x)}$$
$$implies frac{frac{1}{sin x}cos x}{frac{1}{tan x}sec^2 x} = frac{tan x}{sin x}cdot frac{cos x}{sec^2 x} = frac{1}{cos x}cdot cos^3 x = cos^2 x$$
Now, what does $cos^2 x$ tend to as $x$ tends to $0$?
answered Nov 16 at 11:59
KM101
1,823313
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
HINT
As noticed we need $x>0$ then
$$frac{ln sin(x)}{ln tan(x)}=frac{ln frac{sin(x)}x+ln x}{ln frac{tan(x)}x+ln x}$$
answered Nov 16 at 12:07
gimusi
85.5k74294
add a comment |
up vote
1
down vote
HINT
As noticed we need $x>0$ then
$$frac{ln sin(x)}{ln tan(x)}=frac{ln frac{sin(x)}x+ln x}{ln frac{tan(x)}x+ln x}$$
answered Nov 16 at 12:07
gimusi
85.5k74294
add a comment |
up vote
1
down vote
up vote
1
down vote
HINT
As noticed we need $x>0$ then
$$frac{ln sin(x)}{ln tan(x)}=frac{ln frac{sin(x)}x+ln x}{ln frac{tan(x)}x+ln x}$$
answered Nov 16 at 12:07
gimusi
85.5k74294
HINT
As noticed we need $x>0$ then
$$frac{ln sin(x)}{ln tan(x)}=frac{ln frac{sin(x)}x+ln x}{ln frac{tan(x)}x+ln x}$$
answered Nov 16 at 12:07
gimusi
85.5k74294
answered Nov 16 at 12:07
gimusi
85.5k74294
answered Nov 16 at 12:07
gimusi
85.5k74294
answered Nov 16 at 12:07
gimusi
85.5k74294
85.5k74294
add a comment |
add a comment |
up vote
1
down vote
Just consider the reciprocal (for $x>0$):
$$frac{ln tan x}{ln sin x} = frac{ln sin x - ln cos x}{ln sin x} =1- frac{ln cos x}{ln sin x} stackrel{x to 0^+}{longrightarrow} 1$$
It follows $boxed{lim_{x to 0+} frac{ln sin(x)}{ln tan(x)} = 1}$.
answered Nov 16 at 12:17
trancelocation
8,0561519
add a comment |
up vote
1
down vote
Just consider the reciprocal (for $x>0$):
$$frac{ln tan x}{ln sin x} = frac{ln sin x - ln cos x}{ln sin x} =1- frac{ln cos x}{ln sin x} stackrel{x to 0^+}{longrightarrow} 1$$
It follows $boxed{lim_{x to 0+} frac{ln sin(x)}{ln tan(x)} = 1}$.
answered Nov 16 at 12:17
trancelocation
8,0561519
add a comment |
up vote
1
down vote
up vote
1
down vote
Just consider the reciprocal (for $x>0$):
$$frac{ln tan x}{ln sin x} = frac{ln sin x - ln cos x}{ln sin x} =1- frac{ln cos x}{ln sin x} stackrel{x to 0^+}{longrightarrow} 1$$
It follows $boxed{lim_{x to 0+} frac{ln sin(x)}{ln tan(x)} = 1}$.
answered Nov 16 at 12:17
trancelocation
8,0561519
Just consider the reciprocal (for $x>0$):
$$frac{ln tan x}{ln sin x} = frac{ln sin x - ln cos x}{ln sin x} =1- frac{ln cos x}{ln sin x} stackrel{x to 0^+}{longrightarrow} 1$$
It follows $boxed{lim_{x to 0+} frac{ln sin(x)}{ln tan(x)} = 1}$.
answered Nov 16 at 12:17
trancelocation
8,0561519
edited Nov 16 at 13:17
answered Nov 16 at 12:17
trancelocation
8,0561519
answered Nov 16 at 12:17
trancelocation
8,0561519
answered Nov 16 at 12:17
trancelocation
8,0561519
8,0561519
add a comment |
add a comment |
up vote
0
down vote
The domain of $ln x$ includes all positive values of $x$, so only the right-hand limit exists. If you want to find that, you can use L’Hôpital’s Rule.
$$lim_{x to 0} frac{ln(sin x)}{ln(tan x)}$$
$$implies frac{frac{1}{sin x}cos x}{frac{1}{tan x}sec^2 x} = frac{tan x}{sin x}cdot frac{cos x}{sec^2 x} = frac{1}{cos x}cdot cos^3 x = cos^2 x$$
Now, what does $cos^2 x$ tend to as $x$ tends to $0$?
answered Nov 16 at 11:59
KM101
1,823313
add a comment |
up vote
0
down vote
The domain of $ln x$ includes all positive values of $x$, so only the right-hand limit exists. If you want to find that, you can use L’Hôpital’s Rule.
$$lim_{x to 0} frac{ln(sin x)}{ln(tan x)}$$
$$implies frac{frac{1}{sin x}cos x}{frac{1}{tan x}sec^2 x} = frac{tan x}{sin x}cdot frac{cos x}{sec^2 x} = frac{1}{cos x}cdot cos^3 x = cos^2 x$$
Now, what does $cos^2 x$ tend to as $x$ tends to $0$?
answered Nov 16 at 11:59
KM101
1,823313
add a comment |
up vote
0
down vote
up vote
0
down vote
The domain of $ln x$ includes all positive values of $x$, so only the right-hand limit exists. If you want to find that, you can use L’Hôpital’s Rule.
$$lim_{x to 0} frac{ln(sin x)}{ln(tan x)}$$
$$implies frac{frac{1}{sin x}cos x}{frac{1}{tan x}sec^2 x} = frac{tan x}{sin x}cdot frac{cos x}{sec^2 x} = frac{1}{cos x}cdot cos^3 x = cos^2 x$$
Now, what does $cos^2 x$ tend to as $x$ tends to $0$?
answered Nov 16 at 11:59
KM101
1,823313
The domain of $ln x$ includes all positive values of $x$, so only the right-hand limit exists. If you want to find that, you can use L’Hôpital’s Rule.
$$lim_{x to 0} frac{ln(sin x)}{ln(tan x)}$$
$$implies frac{frac{1}{sin x}cos x}{frac{1}{tan x}sec^2 x} = frac{tan x}{sin x}cdot frac{cos x}{sec^2 x} = frac{1}{cos x}cdot cos^3 x = cos^2 x$$
Now, what does $cos^2 x$ tend to as $x$ tends to $0$?
answered Nov 16 at 11:59
KM101
1,823313
answered Nov 16 at 11:59
KM101
1,823313
answered Nov 16 at 11:59
KM101
1,823313
answered Nov 16 at 11:59
KM101
1,823313
1,823313
add a comment |
add a comment |
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Hint:As can be seen, the point is not in the domain of the function, so the limit does not exist.
– 1ENİGMA1
Nov 16 at 11:50
You can only take right hand limit. To find the value of the limit you simply have to apply L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 11:52
1
@1ENİGMA1 The existence of limits does not depend on whether the function is defined at the point. We only require that the point is a limit point of the domain.
– xbh
Nov 16 at 11:56
@xbh I know it. What I mean you can see answers.
– 1ENİGMA1
Nov 16 at 12:16